Abstract
Let G be a non-abelian finite simple group. In addition, let \(\Delta _G\) be the intersection graph of G, whose vertices are the proper non-trivial subgroups of G, with distinct subgroups joined by an edge if and only if they intersect non-trivially. We prove that the diameter of \(\Delta _G\) has a tight upper bound of 5, thereby resolving a question posed by Shen (Czechoslov Math J 60(4):945–950, 2010). Furthermore, a diameter of 5 is achieved only by the baby monster group and certain unitary groups of odd prime dimension.
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1 Introduction
For a finite group G, let \(\Delta _G\) be the intersection graph of G. This is the graph whose vertices are the proper non-trivial subgroups of G, with two distinct vertices \(S_1\) and \(S_2\) joined by an edge if and only if \(S_1 \cap S_2 \ne 1\). We write \(d(S_1,S_2)\) to denote the distance in \(\Delta _G\) between vertices \(S_1\) and \(S_2\), and if these vertices are joined by an edge, then we write \(S_1 \sim S_2\). Additionally, \(\mathrm {diam}(\Delta _G)\) denotes the diameter of \(\Delta _G\).
Csákány and Pollák [5] introduced the graph \(\Delta _G\) in 1969 as an analogue of the intersection graph of a semigroup defined by Bosák [1] in 1964. For finite non-simple groups G, Csákány and Pollák determined the cases where \(\Delta _G\) is connected, and proved that, in these cases, \(\mathrm {diam}(\Delta _G) \leqslant 4\) (see also [14, Lemma 5]). It is not known if there exists a finite non-simple group G with \(\mathrm {diam}(\Delta _G) = 4\).
Suppose now that G is a non-abelian finite simple group. In 2010, Shen [14] proved that \(\Delta _G\) is connected, and asked two questions: does \(\mathrm {diam}(\Delta _G)\) have an upper bound? If yes, does the upper bound of 4 from the non-simple case also apply here? In the same year, Herzog, Longobardi, and Maj [7] independently showed that the subgraph of \(\Delta _G\) induced by the maximal subgroups of G is connected with diameter at most 62. As each proper non-trivial subgroup of G is adjacent in \(\Delta _G\) to some maximal subgroup, this implies an upper bound of 64 for \(\mathrm {diam}(\Delta _G)\), resolving Shen’s first question. Ma [12] reduced this upper bound to 28 in 2016. In the other direction, Shahsavari and Khosravi [13, Theorem 3.7] proved in 2017 that \(\mathrm {diam}(\Delta _G) \geqslant 3\).
In this paper, we significantly reduce the previously known upper bound of 28 for \(\mathrm {diam}(\Delta _G)\), and show that the new bound is best possible. In particular, we prove the following theorem, which resolves Shen’s second question with a negative answer.
Theorem 1.1
Let G be a non-abelian finite simple group.
-
(i)
\(\Delta _G\) is connected with diameter at most 5.
-
(ii)
If G is the baby monster group \({\mathbb {B}}\), then \(\mathrm {diam}(\Delta _G) = 5\).
-
(iii)
If \(\mathrm {diam}(\Delta _G) = 5\) and \(G \not \cong {\mathbb {B}}\), then G is a unitary group \(\mathrm {U}_n(q)\), with n an odd prime and q a prime power.
Remark 1.2
Using information from the Atlas [4], we can show that if \(S_1\) and \(S_2\) are vertices of \(\Delta _{{\mathbb {B}}}\) with \(d(S_1,S_2) = 5\), then \(|S_1| = |S_2| = 47\).
Remark 1.3
If \(G \in \{\mathrm {U}_3(3), \mathrm {U}_3(5),\mathrm {U}_5(2)\}\), then G has no maximal subgroup of odd order [11, Theorem 2]. As we will explain in the proof of Theorem 1.1, this implies that \(\mathrm {diam}(\Delta _G) \leqslant 4\). Indeed, we can use information from the Atlas [4] to show that \(\mathrm {diam}(\Delta _{\mathrm {U}_3(3)}) = 3\). Furthermore, even though \(\mathrm {U}_3(7)\) has a maximal subgroup of odd order, we deduce from calculations in Magma [2] that \(\mathrm {diam}(\Delta _{\mathrm {U}_3(7)}) = 4\). On the other hand, we can adapt the proof of Theorem 1.1(ii), with the aid of several Magma calculations, to show that \(\mathrm {diam}(\Delta _{\mathrm {U}_7(2)}) = 5\).
It is an open problem to classify the finite simple unitary groups G with \(\mathrm {diam}(\Delta _G) = 5\).
2 Proof of Theorem 1.1
In order to prove Theorem 1.1 in the unitary case, we will require the following proposition. For a prime power q, let f be the unitary form on the vector space \(V:={\mathbb {F}}_{\!q^2}^3\) whose Gram matrix is the \(3 \times 3\) identity matrix, and let \(\mathrm {SU}_3(q)\) be the associated special unitary group. Then the standard basis for (V, f) is orthonormal, and a matrix \(A \in \mathrm {SL}_3(q^2)\) lies in \(\mathrm {SU}_3(q)\) if and only if \(A^{-1} = A^{\sigma {\mathsf {T}}}\), where \(\sigma \) is the field automorphism \(\alpha \mapsto \alpha ^q\) of \({\mathbb {F}}_{\!q^2}\). For a subspace U of V, we will write \(\mathrm {SU}_3(q)_U\) to denote the stabiliser of U in \(\mathrm {SU}_3(q)\).
Proposition 2.1
Let q be a prime power greater than 2, and let X and Y be one-dimensional subspaces of the unitary space (V, f), with X non-degenerate. Then \(\mathrm {SU}_3(q)_X \cap \mathrm {SU}_3(q)_Y\) contains a non-scalar matrix.
Proof
We may assume without loss of generality that X contains the vector (1, 0, 0). Let (a, b, c) be a non-zero vector of Y. In addition, let \(\omega \) be a primitive element of \({\mathbb {F}}_{\!q^2}\), and let \(\lambda :=\omega ^{q-1}\). Then \(|\lambda | = q+1 > 3\). If at least one of a, b, and c is equal to 0, then \(\mathrm {SU}_3(q)_X \cap \mathrm {SU}_3(q)_Y\) contains a non-scalar diagonal matrix with two diagonal entries equal to \(\lambda \) and one equal to \(\lambda ^{-2}\) (not necessarily in that order).
Suppose now that a, b, and c are all non-zero, and let \(\mu :=b^{-1}c\). We may assume that \(a = 1\). The trace map \({\alpha \mapsto \alpha + \alpha ^q}\) from \({\mathbb {F}}_{\!q^2}\) to \({\mathbb {F}}_{\!q}\) is \({\mathbb {F}}_{\!q}\)-linear, and hence has a non-trivial kernel. In particular, there exists \(\beta \in {\mathbb {F}}_{\!q^2}\) such that \(\beta \ne 1\) and \(\beta + \beta ^q = 2\). It follows from simple calculations that if \(\mu ^{q+1} = -1\), then \(\mathrm {SU}_3(q)_X \cap \mathrm {SU}_3(q)_Y\) contains
If instead \(\mu ^{q+1} \ne -1\), then we can define \(\gamma :=\lambda ^{-2}(\lambda ^3+\mu ^{q+1})(1+\mu ^{q+1})^{-1}\). In this case, \(\mathrm {SU}_3(q)_X \cap \mathrm {SU}_3(q)_Y\) contains
Note that \(\lambda \ne \gamma \) since \(|\lambda | > 3\). \(\square \)
Proof of Theorem 1.1
Let \(S_1\) and \(S_2\) be proper non-trivial subgroups of G, and let \(M_1\) and \(M_2\) be maximal subgroups of G that contain \(S_1\) and \(S_2\), respectively. Since \(d(M_1,M_2) \leqslant d(S_1,M_2) \leqslant d(S_1,S_2)\), we may assume that \(S_1\) and \(S_2\) are not maximal in G. We may also assume that \(M_1 \ne M_2\), as otherwise \(S_1 \sim M_1 \sim S_2\) and \(d(S_1,S_2) \leqslant 2\).
Suppose first that \(|M_1|\) and \(|M_2|\) are even. Then, as observed in the proof of [7, Proposition 3.1], there exist involutions \(x \in M_1\) and \(y \in M_2\), with \(\langle x,y \rangle \) equal to a (proper) dihedral subgroup D of G (with \(|D| = 2\) allowed). Hence \(S_1 \sim M_1 \sim D \sim M_2 \sim S_2\), and so \(d(S_1,S_2) \leqslant 4\). In particular, if every maximal subgroup of G has even order, then \(\mathrm {diam}(\Delta _G) \leqslant 4\), as noted in the proof of [12, Lemma 2.3].
It remains to consider the case where G contains a maximal subgroup of odd order. Liebeck and Saxl [11, Theorem 2] present a list containing all possibilities for G and its maximal subgroups of odd order. By the previous paragraph, we may assume that the maximal subgroup \(M_1\) has odd order. However, \(|M_2|\) may be even. In what follows, information about the sporadic simple groups is taken from the Atlas [4], except where specified otherwise.
(i) \(G = A_p\), with p prime, \(p \equiv 3 \pmod 4\), and \(p \notin \{7,11,23\}\). By [5, Theorem 2] (see also [14, Assertion I]), the intersection graph of any simple alternating group has diameter at most 4.
(ii) \(G = \mathrm {L}_2(q)\), with q a prime power and \(q \equiv 3 \pmod 4\). The group G acts transitively on the set \(\Omega \) of one-dimensional subspaces of the vector space \({\mathbb {F}}_{\!q}^2\). Additionally, \(M_1 = G_U\) for some \(U \in \Omega \), and \(G_U \cap G_W \ne 1\) for each \(W \in \Omega \). If \(|M_2|\) is odd, then \(M_2 = G_W\) for some W, and it follows that \(M_1 \sim M_2\) and \(d(S_1,S_2) \leqslant 3\). We may therefore assume that \(M_2\) contains an involution g. Then g fixes no subspace in \(\Omega \), and so \(g \in G_{\{U,X\}}\) for some \(X \in \Omega \setminus \{U\}\). Since the non-trivial subgroup \(G_U \cap G_X\) lies in both \(M_1 = G_U\) and \(G_{\{U,X\}}\), we deduce that \(S_1 \sim M_1 \sim G_{\{U,X\}} \sim M_2 \sim S_2\). Thus \(d(S_1,S_2) \leqslant 4\).
(iii) \(G = \mathrm {L}_n(q)\), with n an odd prime, q a prime power, and \(G \not \cong \mathrm {L}_3(4)\). Similarly to the previous case, the group G and its overgroup \(R:=\mathrm {PGL}_n(q)\) act transitively on the set \(\Omega \) of one-dimensional subspaces of the vector space \({\mathbb {F}}_{\!q}^n\). Here, \(M_1 = {G \cap N_R(K)}\), where K is a Singer subgroup of R, i.e., a cyclic subgroup of order \((q^n-1)/(q-1)\) (see [8, §1–2]).
Now, \(M_1\) contains a non-identity element m that fixes a subspace \(X \in \Omega \) [8, p. 497]. Observe that \({m^k \in M_1}\) for each \(k \in K\). The action of K on \(\Omega \) is transitive, and hence each subspace in \(\Omega \) is fixed by some non-identity element of \(M_1\). Therefore, if a non-identity element of \(S_2\) fixes a subspace \(U \in \Omega \), then \(S_1 \sim M_1 \sim G_U \sim S_2\) and \(d(S_1,S_2) \leqslant 3\). Otherwise, since n is prime, there exists \(g \in G\) such that \(S_2 \cap M_1^g \ne 1\). Thus \({S_1 \sim M_1 \sim G_X \sim M_1^g \sim S_2}\) and \(d(S_1,S_2) \leqslant 4\).
(iv) \(G = \mathrm {U}_n(q)\), with n an odd prime, q a prime power, and \(G \not \cong \mathrm {U}_3(3)\), \(\mathrm {U}_3(5)\), or \(\mathrm {U}_5(2)\). Here, G acts intransitively on the set of one-dimensional subspaces of the vector space \({\mathbb {F}}_{\!q^2}^n\). Let \((q+1,n)\) denote the greatest common divisor of \(q+1\) and n. The maximal subgroup \(M_1\) is equal to \(N_G(T)\), where T is a Singer subgroup of G, i.e., a cyclic subgroup of order \(\frac{q^n+1}{(q+1)(q+1,n)}\) (see [8, §5]). In fact, each maximal subgroup of G of odd order is conjugate to \(M_1\). Similarly to the linear case, \(M_1\) contains a non-identity element that fixes a one-dimensional subspace X of \({\mathbb {F}}_{\!q^2}^n\) [8, p. 512].
Let \(L:=G_X\). Then \(M_1 \sim L\), and we can calculate |L| using [3, Table 2.3]. In particular, |L| is even. Hence if \(|M_2|\) is even, then G contains a dihedral subgroup D such that \({S_1 \sim M_1 \sim L \sim D \sim M_2 \sim S_2}\), and \(d(S_1,S_2) \leqslant 5\). If \(|M_2|\) is odd, then there exists an element \(g \in G\) such that \(M_2 = M_1^g\). Thus \(L^g \sim M_2\). If \(n = 3\) and X is non-degenerate, then it follows from Proposition 2.1 that \(L \sim L^g\). Therefore, \({S_1 \sim M_1 \sim L \sim L^g \sim M_2 \sim S_2}\) and \(d(S_1,S_2) \leqslant 5\). In the remaining cases, we will show that \(|L|^2/|G| > 1\), and hence \(|L|\,|L^g| > |G|\). It will follow that \(L \cap L^g \ne 1\), again yielding \(d(S_1,S_2) \leqslant 5\).
Observe that \(|L|^2/|G| > 1\) if and only if \(\log |G|/\log |G:L| > 2\). By [6, Proposition 3.2], if \(n \geqslant 7\), then \(\log |G|/\log |G:L| > 2\), as required. If instead \(n = 3\), then we may assume that X is totally singular. Here, \(q > 2\), and hence
Suppose finally that \(n = 5\). If X is totally singular, then \(|L|^2/|G|\) is equal to
If instead X is non-degenerate, then
(v) \(G = \mathrm {M}_{23}\). In this case, \(M_1\) has shape \(23 :11\). We argue as in the proof of [14, Assertion I]. There exists a maximal subgroup L of G isomorphic to \(\mathrm {M}_{22}\), and \(|M_1|\,|L|\) and \(|M_2|\,|L|\) are greater than |G| (for any choice of \(M_2\)). It follows that \(S_1 \sim M_1 \sim L \sim M_2 \sim S_2\), and so \(d(S_1,S_2) \leqslant 4\).
(vi) \(G = \mathrm {Th}\). Here, \(M_1\) has shape \(31 :15\). If the proper non-trivial subgroup \(S_1\) of \(M_1\) has order 31, then \(S_1\) lies in a maximal subgroup of shape \(2^5\cdot \mathrm {L}_5(2)\). Otherwise, \(|C_G(S_1)|\) is even. Therefore, in each case, \(S_1\) lies in a maximal subgroup of even order. The same is true for \(S_2\), and thus \(d(S_1,S_2) \leqslant 4\).
(vii) \(G = {\mathbb {B}}\). In this case, \(M_1\) has shape \(47 :23\). Additionally, G has a maximal subgroup \(K \cong \mathrm {Fi}_{23}\), which has even order, and \(M_1 \sim K\). Hence if \(|M_2|\) is even, then \(S_1 \sim M_1 \sim K \sim D \sim M_2 \sim S_2\) for some dihedral subgroup D of G, yielding \(d(S_1,S_2) \leqslant 5\). Otherwise, there exists an element \(g \in G\) such that \(M_2 = M_1^g\), and hence \(K^g \sim M_2\). As \(|K|^2/|G| > 1\), we conclude that \({S_1 \sim M_1 \sim K \sim K^g \sim M_2 \sim S_2}\) and \(d(S_1,S_2) \leqslant 5\). Thus \(\mathrm {diam}(\Delta _G) \leqslant 5\).
We now show that \(\mathrm {diam}(\Delta _G)\) is equal to 5. Let H be a subgroup of \(M_1\) of order 23. Then H is a Sylow subgroup of G. It follows from [15, p. 67] that each maximal subgroup of G that contains H is conjugate either to \(M_1\), to K, or to a subgroup L of shape \(2^{1+22}\cdot \mathrm {Co}_2\). We may assume that \(H \leqslant M_1 \cap K \cap L\). Additionally, \(N_G(H)\) has shape \((23 :11) \times 2\) and \(N_L(H) = N_G(H)\), while \({|N_G(H):N_{M_1}(H)|} = 22\). Since the 22 non-identity elements of H fall into two K-conjugacy classes and \(C_K(H) = H\), we conclude that \(N_K(H)\) has shape \(23 :11\), and so \({|N_G(H):N_K(H)|} = 2\).
Consider the pairs \((H',M')\), where \(H'\) is a G-conjugate of H, \(M'\) is a G-conjugate of \(M_1\), and \(H' \leqslant M'\). As any two G-conjugates of H appear in an equal number of such pairs, we deduce that H lies in exactly \({|N_G(H):N_{M_1}(H)|} = 22\) G-conjugates of \(M_1\). Similarly, H lies in two G-conjugates of K and one G-conjugate of L.
As \(M_1\) has shape \(47 :23\), it contains a subgroup S of order 47. In fact, \(M_1\) is the unique maximal subgroup of G that contains S. Hence if J is a maximal subgroup of G satisfying \(J \ne M_1\) and \(J \cap M_1 \ne 1\), then J contains a G-conjugate of H. Let \({\mathcal {U}}\) be the set of G-conjugates of H that lie in at least one such maximal subgroup J, or in \(M_1\). There are 47 subgroups of order 23 in \(M_1\), each of which lies in two G-conjugates of K, and there are \(|K:N_K(H)|\) subgroups of order 23 in K. Therefore, there are fewer than \(47 \cdot 2|K:N_K(H)|\) subgroups in \({\mathcal {U}}\) that lie in at least one G-conjugate of K. By considering the G-conjugates of \(M_1\) and L similarly, we conclude that
Hence there exists \(g \in G\) such that no subgroup of \(M_1^g\) lies in \({\mathcal {U}}\). This means that \(M_1\) and \(M_1^g\) are not adjacent in \(\Delta _G\) and have no common neighbours, and so \(d(M_1,M_1^g) > 2\). As \(M_1\) and \(M_1^g\) are the unique neighbours of S and \(S^g\), respectively, it follows that \(d(S,S^g) > 4\). Therefore, \(\mathrm {diam}(\Delta _G) = 5\).
(viii) \(G = {\mathbb {M}}\). Liebeck and Saxl list two possible maximal subgroups of odd order (up to conjugacy), of shape \(59 :29\) and \(71 :35\), respectively. However, these subgroups are not, in fact, maximal: the former lies in the maximal subgroup \(\mathrm {L}_2(59)\) constructed in [9], and the latter lies in the maximal subgroup \(\mathrm {L}_2(71)\) constructed in [10]. Hence G has no maximal subgroup of odd order, and so \(\mathrm {diam}(\Delta _G) \leqslant 4\). \(\square \)
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Acknowledgements
The author is grateful to Colva Roney-Dougal and Peter Cameron for proofreading this paper and providing helpful feedback, and to Peter for the linear group arguments used in the proof of Theorem 1.1.
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Freedman, S.D. The intersection graph of a finite simple group has diameter at most 5. Arch. Math. 117, 1–7 (2021). https://doi.org/10.1007/s00013-021-01583-3
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DOI: https://doi.org/10.1007/s00013-021-01583-3